Problem: If s=βββ1+2β321ββ βββββ1+2β161ββ βββββ1+2β81ββ βββββ1+2β41ββ βββββ1+2β21ββ ββ, then s is equal to
Answer Choices:
A. 21ββββ1β2β321ββ βββ1
B. βββ1β2β321ββ βββ1
C. 1β2β321β
D. 21ββββ1β2β321ββ ββ
E. 21β
Solution:
Let x=2β321β. Then s=(1+x)(1+x2)(1+x4)(1+x8)(1+x16)
Now (1βx)s=1βx32=21β. β΄s=21β(1βx)β1=21ββββ1β2β321ββ βββ1